variance of a random variable on unit interval with mean inequality Let $X$ be a random variable restricted to the unit interval (ie for probability space $\Omega$, $X:\Omega \rightarrow [0,1]$).
By the Popoviciu Inequality, I have an upper bound on the variance of $X$ which evaluates to var$(X) \leq \frac{1}{4}$.
I further know how to find a constant $0<a\in\mathbb{R}$ as a lower bound on the mean yielding $ 0<a\leq E[X]  $.  Obviously, the restriction of $X$ to the unit interval trivially yields $E[X]\leq1$ by the Edmundson-Madansky Inequality.
Using the above information, is there any sharper bound I can place on var$(X)$?
 A: I will give a solution in the spirit of Bertsimas, D., & Popescu, I. (2005). Optimal inequalities in probability theory: A convex optimization approach. SIAM Journal on Optimization, 15(3), 780-804.
 A more compact exposition of the approach is also available in section 7.4 of Convex Optimization book by Stephen Boyd and Lieven Vandenberghe.
Consider a restricted problem where $E[X]=\bar{a}$. Then the sharpest upper bound on the variance $Var(X)=E[X^2]-(E[X])^2=E[X^2]-\bar{a}^2$ is obtained from the moment problem (similarly to (2.1) in the cited article)
\begin{align}
\max&\ E[X^2]\\
\mbox{s.t.}&\ E[X]&=\bar{a},\\
&\ E[1\{X\in[0,1]\}]&=1,
\end{align}
where the maximization is over all possible random variables with $[0,1]$ support set. The dual to this problem is (similarly to (2.2) in the cited article)
\begin{align}
\min&\ \bar{a}\lambda+\mu\\
\mbox{s.t.}&\ x\lambda +\mu\ge x^2,\quad \forall x\in[0,1],
\end{align}
where the minimization is over $\lambda$ and $\mu$ -- Lagrange multipliers of the constraints. Since $x^2-\lambda x-\mu$ is convex, it attains maximum at the boundaries of the interval (i.e., 0 and 1), so the dual is equivalent to 
\begin{align}
\min&\ \ \bar{a}\lambda+\mu\\
\mbox{s.t.}
&\ \ \lambda +\mu\ge 1,\\
&\ \ \mu\ge 0.
\end{align}
Since $\bar{a}\le 1,$ this LP has optimal value of $\bar{a}$ attained at $\mu=0,\lambda=1.$ You are now left to maximize $Var(X)=\bar{a}-\bar{a}^2$ subject to $\bar{a}\ge a.$ The maximum is either at $\bar{a}=\frac12$ if $a\le\frac12$ or at $\bar{a}=a$ otherwise, i.e. $$Var(x)\le\max\{1/2,a\}-(\max\{1/2,a\})^2\le\frac14.$$
